Let y = x^(a/b) for integers "a" and "b". Raise both sides to the "bth" power and use implicit differentiation to prove the power rule y' = (a/b)x^(a/b-1).

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- October 14th 2008, 01:54 PMerimat89Use implicit differentiation to prove the power rule
Let y = x^(a/b) for integers "a" and "b". Raise both sides to the "bth" power and use implicit differentiation to prove the power rule y' = (a/b)x^(a/b-1).

- October 14th 2008, 02:04 PMvincisonfireReply
y^b=x^a

b*y^(b-1)*y'=a*x^(a-1)

y'={a*x^(a-1)}/{b*y^(b-1)}

y'=(a/b)*x^(a-1)/x^(a*(b-1)/b)

y' = (a/b)x^(a/b-1) - October 14th 2008, 02:32 PMerimat89Confused
What exactly happened to the b*y in the 3rd step to the 4th step? Why did the y variable disappear?

- October 14th 2008, 02:34 PMSoroban
Hello, erimat89!

I'll do it in LaTex . . .

Quote:

Let for integers

Raise both sides to the power and use implicit differentiation

to prove the power rule: .

We have: . .[1]

Raise to the power: . .[2]

Differentiate implicitly: . .[3]

Divide [2] by

Substitute [1]: .

Substitute into [3]: .

- October 14th 2008, 02:34 PMvincisonfireReply
b goes in the (a/b)

y = x^(a/b) that why it disappear - October 14th 2008, 02:47 PMerimat89Thanks
Thanks for your help guys I'm finding the calculus concepts to be relatively easy but I obviously haven't mastered the algebra skills that are required to solve calculus related problems which is the reason I'm having such a hard time.